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# AQA Physics Unit 1 PHYA1 20th May 2013 watch

1. (Original post by Jimmy20002012)
When it said shows how the reaction is conserved with that feynam diagram what did people put I did:

Charge and lepton number and the showed it by writing the equation of:

P ----> n + e+ + neutrino

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The equation had to include up and down quarks, as opposed to protons and neutrons, I'm afraid.
2. Do the answers to these answers to the last page sound familiar to people.

1.03x10^-4
V=0.51
30,000 ohms. ?
3. Am I the only one that found it hard?
4. (Original post by StalkeR47)
Interaction is always either w+ or w - in decay. Proton ---- turns into a neutron. So a hadron is responsible for weak since they decay through weak interaction. Not interact.
You're wrong. A decay is still an interaction. All fermions, that is, all particles with half-integer spin feel the weak force. You probably can't even explain how leptons even "interact" via the weak force so I don't know why you argue that so adamantly. Quarks can undergo neutral current interactions for the transfer of momentum, which are interactions not decays, and are mediated by the Z boson (a weak force mediator).
5. (Original post by Me123456789)
Am I the only one that found it hard?
I found it midway compared to other papers i've done.
6. Hi guys,

The test was okay; didn't expect that 6-marker!

These are my answers. I'm not sure if they are right.

Q1
28 n, 20 p, 18 e
charge=+3.2x10^-19
specific charge=4.0x10^6 Ckg^-1

Q2
D quark, W+ boson, positron, electron neutrino
Weak interaction
B was the exchange boson
photon has infinite range, W+ has limited range
Charge conserved, baryon number conserved

Q3
6 marker, hadrons experience strong and weak, leptons weak only. hadrons are protons neutrons, leptons are electrons, neutrinos. both experience weak, both have antiparticles, both have charge. hadrons include baryons and mesons etc
proton, antiproton
antiparticles have opposite charge sign

Q4
E depends on f. 1 electron absorbs 1 photon. KE=hf-work function. Max value. KE is less than max if work is done on deeper electrons.
work function=1.8eV
frequency=4.4x10^14 Hz
frequency reduced, less energy, less kinetic energy
intensity increased, more photons per second, more electrons per second.

Q5
peak to peak=128V
peak=64V
RMS=45(.3)V
frequency=100Hz
DC= straight ine at 45V
Oscilloscope, y-gain is same as before, 20Vdiv^-1. Change time base to include 2 cycles. 1 cycle=10ms, 2 cycles=20ms, so 2msdiv^-1.

Q6
6.3V
5.7V across 2 ohm resistor
2.85A across 2 ohm resistor
1.35A across unknown resistor
4.2 ohm resistance
2.86 ohms total resistance
power: Internal resistor=26.5W, 2 ohm resistor=16.2W, other resistor=7.7W
energy conserved: battery power-total power dissipated=0
50.4-26.5-7.7-16.2=0

Q7
6.7x10^-5 A
1/3 V
resistance of LDR decreases, so greater proportion of pd across resistor, reading increases
30000 ohms resistance.

Tell me if there are any issues.
Good Luck!

PS, any ideas for grade boundaries?
7. (Original post by Me123456789)
Am I the only one that found it hard?
(Original post by g.k.galloway)
I found it midway compared to other papers i've done.
I didn't think it was especially difficult, but not the easiest. I think it was more that it was a "different" paper compared to most of the past papers, slightly different style questions. I agree with the "midway" comment though.
8. Anyone get 5.99 for one?...

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9. (Original post by GeneralOJB)
Oh I thought you were talking about the max KE question. Decreasing the frequency would mean the emitted electrons have a lower kinetic energy. If you decreased it enough, then yes, less electrons would be emitted and then eventually zero when it goes below f0.
Did you get these answers on the last page? you seem to know what you were doing.

1.03x10^-4 A
V=0.51
30,000 ohms. ?
10. is 20v per division in the y-gain right or not?
11. (Original post by `God)
is 20v per division in the y-gain right or not?
I put 20V it seems the most sensible, but who knows. Does anybody know if it matters which way around you wrote the positron and electron neutrino for the feynman diagram? And also can you say that photon has infinite range and w boson does not for a mark? Cheers.
12. (Original post by ac3xx)
I didn't think it was especially difficult, but not the easiest. I think it was more that it was a "different" paper compared to most of the past papers, slightly different style questions. I agree with the "midway" comment though.
I think lots of exam boards are changing the style of their questions from Jan 13, I've found it to be the case so far, certainly for maths. We will have to see how biology differs, unit 1 (OCR) tomorrow
13. (Original post by Jimmy20002012)
When it said shows how the reaction is conserved with that feynam diagram what did people put I did:

Charge and lepton number and the showed it by writing the equation of:

P ----> n + e+ + neutrino

Posted from TSR Mobile
Hey Jimmy! how are you and how did you find the exam. For me, it was an easy paper, definitely got an A!
14. (Original post by Maloneyyy)
Yeah that's what I did and got 30000 ohms
Really? Because I thought I got like 7xxxxx something, but I can't even remember :L Here's praying I got it right!
15. (Original post by g.k.galloway)
I think lots of exam boards are changing the style of their questions from Jan 13, I've found it to be the case so far, certainly for maths. We will have to see how biology differs, unit 1 (OCR) tomorrow
Yeah, I've found that as well, especially with my Edexcel M1, question style was fairly different to the past papers. Good luck . I've got History wednesday, only subject I struggle on.
16. Argh, i wrote 30 instead of 30,000 for the last question
17. (Original post by ac3xx)
Yeah, I've found that as well, especially with my Edexcel M1, question style was fairly different to the past papers. Good luck . I've got History wednesday, only subject I struggle on.
Yeh S1 was a little weird, but okay
Good luck! I abandoned History at GCSE, too much writing!!
18. (Original post by frony0)
You're wrong. A decay is still an interaction. All fermions, that is, all particles with half-integer spin feel the weak force. You probably can't even explain how leptons even "interact" via the weak force so I don't know why you argue that so adamantly. Quarks can undergo neutral current interactions for the transfer of momentum, which are interactions not decays, and are mediated by the Z boson (a weak force mediator).
I am correct! This is even written in the revision book.
19. (Original post by `God)
is 20v per division in the y-gain right or not?

8 vertical divisions,
Peak to Peak = 128V
so (128/8) = Y-sensitivity = 16.
20. (Original post by Khenir)

8 vertical divisions,
Peak to Peak = 128V
so (128/8) = Y-sensitivity = 16.
I got same.

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